Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises: 23

Answer

$f'(x)=\dfrac{x^{4}e^{x}+2xe^{2x}}{(x^{2}+e^{x})^{2}}$

Work Step by Step

$f(x)=\dfrac{x^{2}e^{x}}{x^{2}+e^{x}}$ Differentiate using the quotient rule: $f'(x)=\dfrac{(x^{2}+e^{x})(x^{2}e^{x})'-(x^{2}e^{x})(x^{2}+e^{x})'}{(x^{2}+e^{x})^{2}}=...$ Apply the product rule to evaluate $(x^{2}e^{x})'$: $...=\dfrac{(x^{2}+e^{x})[(x^{2})(e^{x})'+(e^{x})(x^{2})']-(x^{2}e^{x})(2x+e^{x})}{(x^{2}+e^{x})^{2}}=...$ $\dfrac{(x^{2}+e^{x})(x^{2}e^{x}+2xe^{x})-(x^{2}e^{x})(2x+e^{x})}{(x^{2}+e^{x})^{2}}=...$ Simplify: $...=\dfrac{x^{4}e^{x}+2x^{3}e^{x}+x^{2}e^{2x}+2xe^{2x}-2x^{3}e^{x}-x^{2}e^{2x}}{(x^{2}+e^{x})^{2}}=...$ $...=\dfrac{x^{4}e^{x}+2xe^{2x}}{(x^{2}+e^{x})^{2}}$
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